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Let $A$ be a size $n$ Jordan matrix with $0$ on its diagonal, that is $$A = J_n(0) = [a_{ij}] = \begin{cases} 1, &j=i+1\\ 0, &\text{elsewhere} \end{cases} $$

What is the Jordan Canonical Form of the classical adjoint of A, $\text{adj} A$?

Can we start with the fact that $A$ is singular and $A (\text{adj} A) = 0_n?$

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    $\begingroup$ If the matrix is in Jordan form and is 0s on its diagonal, the last row of the matrix is all 0 and the matrix is singular, shouldn't it be? $\endgroup$ – RGS Nov 14 '16 at 9:35
  • $\begingroup$ Are you talking about the adjoint? $\endgroup$ – user198504 Nov 14 '16 at 9:38
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    $\begingroup$ I am talking about A. A has a row that is only 0, the bottom one. Isn't it? $\endgroup$ – RGS Nov 14 '16 at 9:38
  • $\begingroup$ The first column is zero $\endgroup$ – user198504 Nov 14 '16 at 9:39
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    $\begingroup$ Yup, and the last row as well. Thus $A $ is not non-singular. $A $ is singular $\endgroup$ – RGS Nov 14 '16 at 9:40
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If you just start computing the classical adjoint for $n=2,3,4...$ you should notice a pattern as to what they look like.

$$adj\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & - 1\\ 0 & 0\end{pmatrix}$$

$$adj\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$

$$ adj \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0\end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 & -1\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\end{pmatrix}$$

Once you prove that this pattern holds, the Jordan Form is straightforward to compute.

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    $\begingroup$ Oh cool. But dont you think that the adjoint should alternated between 1 and -1 as n varies. Also, I think the only nonzero entry will be in the topright corner, not in the buttom left corner. $\endgroup$ – user198504 Nov 16 '16 at 8:44
  • $\begingroup$ You are right about the sign. The bottom left corner is definitely correct though. $\endgroup$ – Ken Duna Nov 16 '16 at 14:08
  • $\begingroup$ Oh wait, I forgot to take the transpose! You were right on both counts! $\endgroup$ – Ken Duna Nov 16 '16 at 14:14

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